As usual, I have a ridiculously specific question about a tiny detail of a tiny calculation. There is a tremendous amount of context, so fair warning. I don't need a solution to this whole problem. I need a solution to one specific part, and it is a minor detail. I'll show what I've got so far, and point out my concern at the end.

Our problem is to use residue calculus to compute: $$\int\limits_{-\infty}^{\infty}\frac{\cos(2x)-1}{x^2}dx.$$

We note that this is equal to the real part of the easier-to-work-with integral: $$\int\limits_{-\infty}^{\infty}\frac{e^{2ix}-1}{x^2}dx,$$ and we use a "hippo's back" contour, formed by the positively-oriented half-circle $|z|\leq R$ in the upper half-plane and the line along the real axis from $-R$ to $R$, interrupted from $-\varepsilon$ to $\varepsilon$ by the negatively oriented half-circle in the upper half-plane of radius $\varepsilon$, so as to 'jump' over the simple pole at $z=0$.

Since this contour contains no poles or singularities, etc., the integral of our integrand around it is zero by Cauchy's Theorem. It is also a simple matter to show that the large half-circle's contribution is zero in the limit $R\to\infty$. So now we are left with the fact that in the limit $\varepsilon\to 0$, we obtain:

$$\int\limits_{-\infty}^{\infty}\frac{e^{2ix}-1}{x^2}dx=\lim\limits_{\varepsilon\to 0}\int\limits_{C_\varepsilon}\frac{e^{2iz}-1}{z^2}dz,$$

where $C_\varepsilon$ is the little negatively-oriented half-circle mentioned above.

Now, I know the trick in which the value of the right-hand side happens to give exactly $\pi i \text{Res}(f,0)$ (that is, half of what the full circle would give), and can use it to show that the result is $-2\pi$.

Okay, thanks for reading this far. Here is my actual question: Without resorting to tricks like the half-residue thing, how can we get this result from the right-hand side? That is, how do we prove rigorously that:

$$\lim\limits_{\varepsilon\to 0}\int\limits_{C_\varepsilon}\frac{e^{2iz}-1}{z^2}dz=-2\pi$$

I have been using the parametrization $z=\varepsilon e^{-it}$ with $t\in[0,\pi]$, where the negative in the exponential deals with the negative orientation, which almost always works in these situations, but not here, since the $\varepsilon$ does not entirely cancel from the denominator then as it usually does in these types of problems.

I have confirmed the result with Wolfram-Alpha using the original problem statement, and also noted a few other things, like that the original integrand can be expressed as $\frac{\sin^2(x)}{x^2}$, but gotten nowhere with that.

In a pinch, I'll use the 'half-residue' trick, but being able to do this simply from basic principles would be nice.


The parametrization works (I'm not sure what you mean about it not fully cancelling out of the denominator). For the negatively oriented half circle, you have $$-\lim_{\epsilon\to 0} \int_0^{\pi}\frac{1}{\epsilon}ie^{-it}(1-e^{2i\epsilon e^{it}})dt$$ and you can expand $(1-e^{2i\epsilon e^{it}}) = -2i\epsilon e^{it}+O(\epsilon^2)$ and get $$-\lim_{\epsilon\to 0}\int_0^\pi (2+O(\epsilon)) dt = -2\pi.$$

  • $\begingroup$ Thanks, I'm on my way to bed but I will look at this closely in the morning. $\endgroup$ – The Count Jan 29 '17 at 3:05
  • $\begingroup$ The expansion is what does it. I've never thought that way about these types of problems before. FYI, I used a negative exponent in my parametrization instead of the negative out front, but this solves my quandry. Thanks again. Accepted. $\endgroup$ – The Count Jan 29 '17 at 19:58

For a given, $a$ we have,

$$\int_{-\infty}^\infty \frac{1-\cos(ax)}{x^2}dx =2\int_0^\infty \frac{1-\cos(ax)}{x^2}dx =-2\int_0^\infty \left(\frac{1}{x}\right)'(1-\cos(ax))dx \\=-2\left[ \frac{1-\cos(ax)}{x}\right]^\infty_0 +2a\int_0^\infty\frac{\sin(ax)}{x}dx = 2a\int_0^\infty\frac{\sin(ax)}{x}dx =\color{red}{\pi |a|}$$

Provided that:

$$\lim_{x\to 0} \frac{1-\cos(ax)}{x} = \lim_{x\to 0} a^2x\frac{1-\cos(ax)}{(ax)^2} =0*\frac12 =0.$$

and $$ \int_0^\infty\frac{\sin(ax)}{x}dx =\overset{u=ax}{=}sign(a)a\int_0^\infty\frac{\sin(x)}{x}dx $$


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